If using the method of completing the square to solve the quadratic equation

The curves y = √x and y=(2-x) and the Cartesian axes form two distinct regions in the first quadrant. Find the volumes of rotati

makkiz [27]

Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

$y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\$

And the second one would be

$y = 2-x \,\,\,\,\,, 1 \leq x \leq 2$ .

If you rotate the first region around the "y" axis you get that

$A_1 = 2\pi \int\limits_{0}^{1} x\sqrt{x} dx = \frac{4\pi}{5} = 2.51$

And if you rotate the second region around the "y" axis you get that

$A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188$

And the sum would be 2.51+4.188 = 6.698

If you revolve just the outer curve you get

If you rotate the first region around the x axis you get that

$A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708$

And if you rotate the second region around the x axis you get that

$A_2 = \pi \int\limits_{1}^{2} (2-x)^2 dx = \frac{\pi}{3} = 1.0472$

And the sum would be 1.5708+1.0472 = 2.618

7 0

3 years ago

Please help me im horrable at math

Masja [62]

1.) First plug in the values for m and A:

$2555 = 139(4) + b$

--> $2555 = 556 + b$

Isolate b: $2555 - 556 = 556 - 556 + b$

b = 1999

So your answer is the first option.

2.) Admission costs $5.25, so even if you don't go on any rides, you will still have to pay $5.25. This means that for this problem, 5.25 will be the b value. If each ride costs $2.50, then this means it is $2.50x for the cost of x rides. The total cost will then be $C = 5.25 + 2.50x$ .

7 0

3 years ago

Por término de temporada varias tiendas liquidan pantalones. Su precio de temporada era de $6. 990. Si quiero comprar tres panta

nlexa [21]

La tienda que ofrezca la mejor oferta por tres pantalones será aquella que tenga el mayor descuento de liquidación posible.

<h3>Análisis contextual de una pregunta incompleta</h3>

En este problema no es posible determinar cuántas tiendas existen ni los descuentos de liquidación ( $x$ ) correspondientes a cada uno y que todas las tiendas venden los pantalones al mismo precio de temporada ( $C$ ).

Las condiciones <em>matemáticas</em> de cada variable es:

$C = 6990$ (1)

$x \in$ $[0, 1)$ (2)

$n = 3$ (3)

La fórmula matemática para el precio final ( $C'$ ), en pesos, es la siguiente:

$C' = 3\cdot (6990)\cdot (1-x)$ (4)

Consideremos el caso de $N$ tiendas, cada una con un descuento de liquidación por cada pantalón y que ese descuento es aplicable a la compra de tres pantalones, entonces la tienda que ofrezca la mejor oferta por tres pantalones será aquella que tenga el mayor descuento de liquidación posible. $\blacksquare$